\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>3\\x>\dfrac{-1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 3\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>3\\x< \dfrac{-1}{2}\end{matrix}\right.\)
Vậy x>3 hoặc x<-1/2
\(\left(x-3\right)\left(x+\dfrac{1}{2}\right)>0\)
TH1:
\(\Rightarrow\left[{}\begin{matrix}x-3>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x>3\\x>-\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x>3\)
TH2:
\(\left[{}\begin{matrix}x-3< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x< 3\\x< -\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x< -\dfrac{1}{2}\)
Vậy...........................................
